Go back to FP and BP discussion

**Example #1:** Calculate the freezing point of a solution of 5.00 g of diphenyl C_{12}H_{10} and 7.50 g of naphthalene, C_{10}H_{8} dissolved in 200.0 g of benzene (fp = 5.5 °C)

**Solution**

There is a tiny curve in this problem, but keep in mind that colligative properties are all about how many particles in solution and nothing else.

1) The key to this problem is to calculate moles of each substance and add then together:

(5.00 g / 154.2 g mol¯^{1}) + (7.50 g / 128.2) = 0.0909 mol

2) Calculate the molality:

0.0909 mol / 0.200 kg = 0.455m

3) Calculate the freezing point depression:

ΔT = i K_{f}mx = (1) (5.12 °C

m¯^{1}) (0.455m) = 2.33 °CThe solution freezes at 5.5 - 2.33 = 3.17 °C

**Example #2:** Vitamin K is involved in the blood clotting mechanism. When 0.500 g is dissolved in 10.0 g of camphor, the freezing point is lowered by 4.43 °C. Calculate the molecular weight of vitamin K.

**Solution**

1) To solve this problem, I'd like to engage in an analysis of the units. We will start with the freezing point depression equation:

ΔT = i K_{f}mReplacing the right side with units gives: ΔT = (°C kg mol¯

^{1}) times (mol kg¯^{1})

Notice that i goes away since it is unitless. Next I will replace mol with g / g mol¯^{1}:

ΔT = (°C kg mol¯^{1}) times (g / g mol¯^{1}kg¯^{1})

2) Now, let's insert numbers in the proper place:

4.43 = (40.) times (0.500 / x 0.010)x is the molecular weight of Vitamin K, 0.500 and 4.43 are from the problem and 0.010 is 10 g of camphor done as kilograms. This becomes:

4.43 = 2000 / x

x = 8860 g / mol

We know this is a reasonable answer since vitamins and proteins have molecular weights in the thousands or even tens of thousands.

**Example #3:** A compound containing only boron, nitrogen, and hydrogen was found to be 40.3% B, 52.2% N, and 7.5% H by mass. When 3.301 g of this compound is dissolved in 50.00 g of benzene, the solution produced freezes at 1.30 °C. The freezing point of pure benzene is 5.48 °C; K_{b} for benzene is 5.12 °C *m*^{-1}. What is the molecular weight of this compound?

a. Determine the molecular weight of the solid.

b. Determine the molecular formula of the solid

c. Determine the mole fraction of the solid in the solution

d. If the density of this solution is 0.8989 g/mL , calculate the molarity of the solution

**Solution:**

1) Determine moles of solute using freezing point depression data:

Δt = K_{f}m4.18 °C = 5.12 °C-kg/mol (x / 0.0500 kg)

4.18 ° = (102.4 °C/mol) (x)

x = 0.04082 mol

2) Use moles and 3.301 g to determine molecular weight (answer to part a):

3.301 g / 0.04082 mol = 80.867 g/mol80.9 g/mol (to 3 sig figs)

3) Use an on-line empirical formula calculator to determine the empirical formula:

BNH_{2}

4) Determine the molecular formula (answer to part b):

the "empirical formula weight" of BNH_{2}is 26.8338 gthe molecular weight divided by the "EFW" yields 3

therefore, the molecular formula is:

BComment: the particular substance in this question is Borazine, often referred to as the "inorganic benzene." Its formula is written (BH)_{3}N_{3}H_{6}_{3}(NH)_{3}.

5) Determine the mole fraction of the solute (answer to part c):

moles of solvent = 50.00 g / 78.1134 g/mol = 0.6401 molχ

_{solute}= 0.04082 mol / (0.04082 mol + 0.6401 mol)χ

_{solute}= 0.060

6) Determine molarity of solution (answer to part d):

i) our solution weighs 53.301 g; calculate its volume:0.8989 g/mL = 53.301 g / xii) calculate the molarity:x = 59.30 mL

0.04082 mol / 0.05930 L = 0.6884 M (to 4 sig figs)

**Example #4:** A 1.60-g sample of a mixture of naphthalene (C_{10}H_{8}) and anthracene (C_{14}H_{10}) is dissolved in 20.0 g benzene (C_{6}H_{6}). The freezing point of the solution is found to be 2.81 °C. What is the composition as mass percent of the sample mixture? (The freezing point of benzene is 5.51 °C and K_{f} is 5.12 °C-kg/mol.)

**Solution path #1:**

1) Determine the fp depression if all 1.6 g were naphthalene:

x = (5.12) [(1.6/128.174)/0.020)]x = 3.1956559 °C

2) Determine the fp depression if all 1.6 g were anthracene:

x = (5.12) [(1.6/178.234)/0.020)]x = 2.29810 °C

3) What combination of naphthalene and anthracene provides a Δt = 2.70 °C?

2.70 = (3.1956559) (x) + (2.29810) (1.6 - x)Some algebra reults in:

x = 0.716 g

You may calculate the anthracene on your own. Don't forget to include the weight of the benzene if you calculate the mass percents.

**Solution path #2:**

Δt = 2.70 °CA and B in the following are placeholders for the molalities.

2.70 = (5.12)(A) + (5.12)(B)

You could also write 2.70 = (5.12)(A + B)

A is as follows: numerator is x over molar mass of naphthalene and the denominator is 0.020 kg

B is as follows: numerator is 1.6 - x over molar mass of anthracene and the denominator is 0.020 kg

Keep in mind that both A and B have a fraction in the numerator.

Be careful with the algebra, if you decide to work it out. Don't forget the benzene, if you do the mass percents.

**Example #5:** A 0.265*m* MgSO_{4} solution has a freezing point of -0.61 °C. What is the van 't Hoff factor for this solution? K_{f} = 1.86 °C/*m*

**Solution:**

Δt = i K_{f}m0.61 °C = (x) (1.86 °C/

m) (0.265m)i = 1.24

The theoretical van 't Hoff factor for MgSO_{4} is 2. Because of our value of 1.24, we must conclude that there is a significant amount of ion-pairing in aqueous solutions of MgSO_{4}.

Is this truly the case or is this just a fake problem?

Look at this page. It presents a different set of data (i = 1.12 at 0.1*m*), but it does show that this issue of ion-pairing in MgSO_{4} is not a made-up one.

Take a look at this. The concept of ion-pairing in MgSO_{4} is very real and is of importance in chemical research, both for geochemical and medical reasons.

**Example #6:** A 0.2436 g sample of an unknown substance was dissovled in 20.0 mL of cyclohexane. The density of cyclohexane is 0.779 g/mL. The freezing point depression was 2.50 °C. Calculate molar mass of the unknown substance.

**Solution:**

1) Determine grams of cyclohexane:

(0.779 g/mL) (20.0 mL) = 15.58 gthis is 0.01558 kg

2) Determine moles of unknown substance present:

Δt = i K_{f}m2.50 °C = (1) (20.2) (x / 0.01558)

x = 0.00192822 mol

Note assumption of i = 1 for the solute. Also, the cryoscopic constant for cyclohexane needed to be looked up.

3) Determine molecular weight:

0.2436 g / 0.00192822 mol = 126.3 g/mol

**Example #7:** A 0.124 *m* trichloracetic acid, CCl_{3}COOH (aq), solution has a freezing point of -0.386 °C. What is the percentage ionization of the acid?

**Solution:**

1) Calculate van 't Hoff factor:

Δt = i K_{b}m0.386 = i (1.86) (0.124)

i = 1.6736

2) Calculate value for [H^{+}]:

CCl_{3}COOH <==> H^{+}+ CCl_{3}COO¯total concentration of all ions in solution equals:

(1.6736) (0.124) = 0.20753mThis is a molality, but we will act as if it a molarity since we will assume the density of the solution is 1.00 g/cm

^{3}, which makes the molarity equal to the molality.0.20753 = (0.124 - x) + x + x

x = 0.08353 M

3) Calculate the percent dissociation:

0.08353 / 0.124 = 67.4%

**Bonus problem:** Calculate the percent dissociation of a 0.124 M solution of trichloroacetic acid (K_{a} = 0.170).

You will need to solve a quadratic equation to solve for the hydrogen ion concentration, since 'ignore the minus x' doesn't work. Use a quadratic solver to get the roots, of which the positive root will be correct (remember, you can't have a negative concentration).The answer is 67%. Best wishes with your chem class. Tell your teacher the ChemTeam says hi.

**Example #8:** Cyclohexanol, C_{6}H_{11}OH, is sometimes used as the solvent in molar mass determination. If 0.235 g of benzoic acid C_{7}H_{6}O_{2}, dissolved in 12.45 g of cyclohexanol, lowered the freezing point of pure cyclohexanol by 6.55 °C, what is the molal freezing point constant of the solvent?

**Solution:**

1) Determine moles of benzoic acid:

0.235 g / 122.1224 g/mol = 0.0019243 mol

2) Substitute into the following equation:

Δt = i K_{b}m6.55 = (1) (x) (0.0019243 mol / 0.01245 kg)

x = 42.4 °C/

m

Note: benzoic acid does not dissociate in cyclohexanol, therefore a van't Hoff factor of 1 is used.

**Example #9:** Someday!